Question: Solve for $z$, $ \dfrac{3}{3z + 9} = \dfrac{10}{4z + 12} - \dfrac{z + 5}{5z + 15} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z + 9$ $4z + 12$ and $5z + 15$ The common denominator is $60z + 180$ To get $60z + 180$ in the denominator of the first term, multiply it by $\frac{20}{20}$ $ \dfrac{3}{3z + 9} \times \dfrac{20}{20} = \dfrac{60}{60z + 180} $ To get $60z + 180$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ \dfrac{10}{4z + 12} \times \dfrac{15}{15} = \dfrac{150}{60z + 180} $ To get $60z + 180$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ -\dfrac{z + 5}{5z + 15} \times \dfrac{12}{12} = -\dfrac{12z + 60}{60z + 180} $ This give us: $ \dfrac{60}{60z + 180} = \dfrac{150}{60z + 180} - \dfrac{12z + 60}{60z + 180} $ If we multiply both sides of the equation by $60z + 180$ , we get: $ 60 = 150 - 12z - 60$ $ 60 = -12z + 90$ $ -30 = -12z $ $ z = \dfrac{5}{2}$